Answer to Question #263081 in Physics for Solomon Ossai

Question

An aircraft shell is fired vertically ward with a muzzle velocity at 100ms.calculate

a=the maximum height it can obtain

b=the time taken to teach the height

c=the instantaneous velocities at the end of 20sec and 50sec

Answer 1

(a) Let's take the upwards as the positive direction. Then, we can find the maximum height from the kinematic equation:

$$v^2=v_0^2+2gh,$$ $$0=v_0^2+2gh,$$ $$h=\dfrac{-v_0^2}{2g}=-\dfrac{(1000\ \dfrac{m}{s})^2}{2\times(-9.8\ \dfrac{m}{s^2})}=51\ km.$$

(b) We can find the time that shell takes to reach the maximum height from the kinematic equation:

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$t=\dfrac{-v_0}{g}=\dfrac{-1000\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=102\ s.$$

(c) We can find the instantaneous velocity at the end of $20\ s$ from the kinematic equation:

$$v=v_0+gt=1000\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\times20\ s=804\ \dfrac{m}{s}.$$

We can find the instantaneous velocity at the end of $50\ s$ from the kinematic equation:

$$v=v_0+gt=1000\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\times50\ s=510\ \dfrac{m}{s}.$$